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Physics Question on Thermal Expansion

The density of water at 20∘C20^{\circ}C is 998 kg/m3998\, kg/m^3 and at 40∘C40^{\circ}C 992 kg/m3992\,kg/m^3 . The coefficient of volume expansion of water is

A

3×10−4 ∘C 3 \times 10^{-4} \, ^ \circ C

B

2×10−4 ∘C 2 \times 10^{-4} \, ^ \circ C

C

6×10−4 ∘C 6 \times 10^{-4} \, ^ \circ C

D

10−4 ∘C 10^{-4} \, ^ \circ C

Answer

3×10−4 ∘C 3 \times 10^{-4} \, ^ \circ C

Explanation

Solution

(a) : As PTs=PT11+γΔT=PT11+γ(T2−T1) P T_s = \frac{ PT_1}{1+ \gamma \Delta T} = \frac{ PT_1}{ 1+ \gamma ( T_2-T_1)}
Here,T1=20∘C,T2=40∘CT_1= 20 ^\circ C, T_2 = 40 ^ \circ C
P20=998kg/m3,P40=992kg/m3P_20 = 998 kg / m^3 , P_40 = 992 kg / m^3
∴    992=9981+γ(40−20)\therefore \, \, \, \, 992 = \frac{998}{ 1+\gamma (40-20)}
992=9981+20γ992= \frac{998}{ 1+ 20 \gamma}
992(1+20γ)=998992(1+20\gamma) = 998
1+20γ=998992−1=69921+ 20 \gamma = \frac{998}{992} -1 = \frac{6}{992}
γ=6992×120=3×10−4/∘C\gamma = \frac{6}{992} \times \frac{1}{20}= 3 \times 10^{-4} / ^\circ C